Two prisms A and B have dispersive powers of 0.012 and 0.018 respectively. The two prisms are in contact with each other. The prism A produces a mean deviation of `1.2^(@)`, the mean deviation produced by B if the combination is achromatic is

Options

A. `3.6^(@)`
B. `0.8^(@)`
C. `0.4^(@)`
D. `1.8^(@)`

Answer

To solve the problem, we need to find the mean deviation produced by prism B when the combination of prisms A and B is achromatic.

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Dispersive power of prism A, \( \omega_A = 0.012 \)
- Dispersive power of prism B, \( \omega_B = 0.018 \)
- Mean deviation produced by prism A, \( \delta_A = 1.2^\circ \)
- Mean deviation produced by prism B, \( \delta_B \) (to be determined)

2. **Condition for Achromatic Combination:**
- For the combination to be achromatic, the total deviation must be zero:
\[ \theta_A + \theta_B = 0 \]
- This implies that the deviations caused by the two prisms must balance each other out.

3. **Relate Deviation to Dispersive Power:**
- The deviation \( \theta \) can be expressed in terms of the dispersive power and mean deviation:
\[ \theta = \omega \cdot \delta \]
- Therefore, for prisms A and B, we have:
\[ \theta_A = \omega_A \cdot \delta_A \]
\[ \theta_B = \omega_B \cdot \delta_B \]

4. **Set Up the Equation:**
- Since the combination is achromatic:
\[ \theta_A + \theta_B = 0 \implies \omega_A \cdot \delta_A + \omega_B \cdot \delta_B = 0 \]
- Rearranging gives:
\[ \omega_A \cdot \delta_A = -\omega_B \cdot \delta_B \]

5. **Substitute Known Values:**
- Plugging in the values:
\[ 0.012 \cdot 1.2 = -0.018 \cdot \delta_B \]

6. **Solve for \( \delta_B \):**
- Calculate \( 0.012 \cdot 1.2 \):
\[ 0.012 \cdot 1.2 = 0.0144 \]
- Now, substituting back:
\[ 0.0144 = 0.018 \cdot \delta_B \]
- Solving for \( \delta_B \):
\[ \delta_B = \frac{0.0144}{0.018} = 0.8^\circ \]

### Conclusion:
The mean deviation produced by prism B, when the combination is achromatic, is \( \delta_B = 0.8^\circ \).